Termination w.r.t. Q proof of /tmp/tmpqZMqZD/thiemann12.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HALF(s(s(x))) → HALF(x)
LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
LOG2(x, y) → LE(x, s(0))
LOG2(x, y) → INC(y)
IF(false, x, y) → LOG2(half(x), y)
IF(false, x, y) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INC(s(x)) → INC(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

s₁ > INC₁

Status:

INC₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = LE(x2)
s(x1) = s(x1)

Lexicographic path order with status [LPO].
Precedence:

s₁ > LE₁

Status:

LE₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HALF(s(s(x))) → HALF(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

s₁ > HALF₁

Status:

HALF₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, x, y) → LOG2(half(x), y)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.